Solutions to Differential Equations with Boundary Conditions

This article provides detailed solutions to differential equations with boundary conditions for regular singular points, Chebyshev's equation, eigenvalues and eigenfunctions, and more. It covers topics such as Chebyshev’s polynomial, eigenvalues and eigenfunctions, and boundary conditions for differential equations.

Q1i)
Given Chebyshev’s equation
(1 – x2)y’’ – xy’ + α2y = 0
A point x0 is said to be a regular singular point of DE y’’ + p(x)y’ + Q(x)y = 0. If the functions p
(x) = (x – x0)P(x) and q(x) = (x – x0)2 Q(x) are both analytic at x0’’
y’’ - x
(1−x2 ) y’ + 2 α
(1−x2 ) y = 0
y’’ - x
( x−1)(1+ x) y’ + 2 α
(x−1)(1+ x) y = 0
here,
P(x) = x
(x−1)(1+x)
Q(x) = 2 α
( x−1)(1+ x)
y’’ –P(x)y’ + Q(x)y = 0
Q1ii)
The Chebyshev’s polynomial o degree n ≥ 0 is defined as
Tn(x) = cos(ncos-1(x)), x ∈ [-1, 1]
Tn(x) = cos(nθ) , where x = cosθ, θ ϵ [0 , π ]
∫
−1
1
( 2 x )
−1
2 ∗T m(x) * Tn(x)dx
= - ∫
π
0
cos ( nθ ) ∗cos ( mθ )∗dθ [ since , dθ= −dx
√1−x2 ]

= ∫
0
π
1
2 [cos {( n+m) θ }¿+ cos {(n−m)θ }]dθ ¿
= 1
2 [ sin ( n+m ) θ
n+ m + sin ( n−m ) θ
n−m ]π
0
= 1
2 [ { sin ( n+m ) π
n+m + sin ( n−m ) π
n−m }−{ sin 0
n+m + sin 0
n−m }
] , m ≠ n
= 1
2 [ ( 0+0 )− ( 0+ 0 ) ]
= 0
Hence ,
∫
−1
1
( 1−x2 )
1
2∗T m(x)*Tn(x)dx = 0, when m ≠ n
Q2)
u’’ + λu = 0
u’(0) = u’(1) = 0
so r2 + λ = 0
r = √ λ− ¿
+¿ ¿ ¿ = i √λ− ¿
+¿ ¿ ¿
so u(n) = c1cos(√ λn) + c2sin(√ λn)
using the first boundary condition u(0) = 0
therefore 0 = c1
So,
u(n) = c2sin(√ λn)
u’(n) = c2√ λ * cos( √ λ n)
using the second boundary condition u’(1) = 0

c2√ λ * cos(1 √ λ)
1 √ λ=−π
2 , 3 π
2
2n+1
2 π , n is an integer
So,
√ λ=2 n+1
2∗1 π
Therefore, λn = ( 2n+ 1 )2
4 π2, n is integer
So, u(n) = c2sin( 2n+ 1
2 πn ¿
For λ = 0, u’’ = 0, u(0) = 0, u’(1) = 0
u(n) = c1 + c2n
u(0) = 0, c1 = 0
u(n) = c2(n)
u’(n) = c2 => u’(1) = 0 => 0 = c2
so, λ = 0, can not be an eigenvalue for this problem
now λ < 0
r = √ λ− ¿
+¿ ¿ ¿ ,
so, u(n) = c1cosh(√ λ n ¿ + c2sinh(√ λ n ¿
u(0) = 0
0 = c1cosh(0 ¿ + c2sinh(0 ¿
0 = c1(1) + c2(0)
c1 = 0
so, u(n) = c2sinh(√−λ n ¿

now, u’(n) = c2√−λ*cosh(√−λ n ¿
for, u’(1) = 0
0 = c2√−λ*cosh(1∗ √− λ ¿
0 = c2√−λ*cosh(√ −λ ¿
cosh(√ −λ ¿ = 0
cosh(n0) ≠ 0, for any n => c2 = 0
since both c1 and c2 = 0, so for λ > 0, there is no eigenvalues
So the eigenvalues are;
Therefore, λn = ( 2n+ 1 )2
4 π2, n is an integer
And the corresponding eigenfunction is
u(x) = c2sin( (2 n+1)
2 πn)
Q3)
u’’ + ku = F(x)
u’(0) = u’(1) = 0
so r2 + k = F(x)
r = √ F ( x ) −k− ¿+¿ ¿ ¿ = i √ F ( x ) −k− ¿+¿ ¿ ¿
so u(n) = c1cos(√ F ( x )−kn) + c2sin(√ F ( x ) −kn)
using the first boundary condition u(0) = 0
therefore 0 = c1
So,
u(n) = c2sin(√ F ( x ) −kn)

u’(n) = c2k * cos( √ F ( x )−k n)
using the second boundary condition u’(1) = 0
c2k * cos(1 √ F ( x ) −k )
1 √F ( x ) −k=−π
2 , 3 π
2
2n+1
2 π , n is an integer
So,
√ F ( x )−k = 2n+1
2∗1 π
Therefore, F(x) -kn = ( 2n+ 1 ) 2
4 π2, n is integer
Whre, F(x) = x
So, u(n) = c2sin( 2n+ 1
2 πn ¿
For k = 0, u’’ = 0, u(0) = 0, u’(1) = 0
u(n) = c1 + c2n
u(0) = 0, c1 = 0
u(n) = c2(n)
u’(n) = c2 => u’(1) = 0 => 0 = c2
so, k = 0, cannot be an eigenvalue for this problem
now k < 0
r = √ F ( x ) −k− ¿+¿ ¿ ¿
so, u(n) = c1cosh(√ F ( x )−k n ¿ + c2sinh(√ F ( x )−k n ¿
u(0) = 0

0 = c1cosh(0 ¿ + c2sinh(0 ¿
0 = c1(1) + c2(0)
c1 = 0
so, u(n) = c2sinh(√−(F ( x )−k )n
now, u’(n) = c2√ −k*cosh(√ −(F ( x ) −k ¿) n ¿
for, u’(1) = 0
0 = c2√−(F ( x )−k ¿) ¿*cosh(1∗ √−( F ( x )−k ¿) ¿
0 = c2√−(F ( x )−k ¿) ¿*cosh(√−(F ( x )−k ¿) ¿
cosh(√ F ( x ) −k= 0
cosh(n0) ≠ 0, for any n => c2 = 0
since both c1 and c2 = 0, so for k > 0, there is no eigenvalues
So the eigenvalues are;
Therefore, x - kn = ( 2n+ 1 ) 2
4 π2, n is an integer
And the corresponding eigen function is
u(x) = c2sin( (2 n+1)
2 πn)
Q4)
(p(x)y’)’ – q(x)y + λr(x)y = 0
y = ρsinθ
py’ = ρ cos θ
θ'= 1
p cos2 θ+ ( λr−q ) sin2 θ
boundary condition are

a1y(0) + a2y’(0) = 0
b1y(1) + b2y’(1) = 0
initial condition are
a1sinθ ( 0 ) + a 2
p(0) cosθ ( 0 )=0
0 ≤ θ ( 0 ) < π
tanθ(1 , λ)= −b 2
b 1 p (1)
θ'= 1
p cos2 θ+ ( λr−q ) sin2 θ
At θ'=0
– 1
p cos2 θ= ( λr−q ) sin2 θ
- 1
p ( λr −q ) = sin2 θ
cos2 θ
- 1
p ( λr −q ) =tan2 θ
( λr −q ) =p tan2 θ
p tan2 θ+ q
r = λ
Hence, shows that θ ( 1 , λ ) is an increasing function of λ